L = illuminance
A = surface
i = intensity
L = i / A ==: i = L * A
i = 6 lux * 4 m^2 = 24 lumen
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
Explanation:
We have,
Semimajor axis is 
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :
G is universal gravitational constant
M is solar mass
Plugging all the values,
Since,
So, the orbital period of a dwarf planet is 138.52 years.
They typically represents different wavelengths of element due to its energy emission in the form of visible light. When an electron of that particular element move from a higher energy level down to a lower energy level, it gives off energy in the form of photon emission. Atom of a certain element has a unique electron arrangement thus it can considered that particular element's spectrum is unique.
